# Uniform Continuity definition

• Oct 16th 2008, 01:32 PM
Uniform Continuity definition
Let $\displaystyle (M,d)$ and $\displaystyle (N,p)$ be metric spaces, and let $\displaystyle f:A \subset M \rightarrow N$. Prove that f is an uniformly continuous mapping on A iff for every pair of sequence $\displaystyle \{ x_k \} , \{ y_k \}$ of A with $\displaystyle d(x_k,y_k) \rightarrow 0$, we have $\displaystyle p(f(x_k),f(y_k)) \rightarrow 0$

Proof.

Suppose that f is uniformly continuous, pick sequences $\displaystyle \{ x_k \} , \{ y_k \}$ of A such that $\displaystyle d( x_k,y_k) \rightarrow 0$

Let $\displaystyle \epsilon > 0$ be given, and let $\displaystyle \delta > 0$, then $\displaystyle \exists K \in \mathbb {N}$ such that whenever $\displaystyle k \geq K$, we have $\displaystyle d(x_k,y_k) < \delta$. Since f is uniformly continuous, we then ahve $\displaystyle p(f(x_k),f(y_k)) < \epsilon$, implies that $\displaystyle p(f(x_k),f(y_k)) \rightarrow 0$

Conversely, suppose that math] \{ x_k \} , \{ y_k \} [/tex] are sequences in A with $\displaystyle d( x_k,y_k) \rightarrow$, we have $\displaystyle p(f(x_k),f(y_k)) \rightarrow 0$.

Let $\displaystyle \epsilon > 0$ be given, pick $\displaystyle \delta > 0$, find $\displaystyle K \in \mathbb {N}$ such that whenever $\displaystyle k \geq K$, we have $\displaystyle d(x_k,y_k) < \delta$.

Now suppose that $\displaystyle x_k \rightarrow x_0$ and $\displaystyle y_k \rightarrow y_0$, then $\displaystyle d(x_0,y_0) < \delta$, but we also have $\displaystyle p(f(x_k),f(y_k)) \rightarrow 0$, which means $\displaystyle p(f(x_k),f(y_k)) < \epsilon$.

Q.E.D.

Is this right? Thanks!
• Oct 16th 2008, 03:21 PM
Plato
Quote:

Let $\displaystyle (M,d)$ and $\displaystyle (N,p)$ be metric spaces, and let $\displaystyle f:A \subset M \rightarrow N$. Prove that f is an uniformly continuous mapping on A iff for every pair of sequence $\displaystyle \{ x_k \} , \{ y_k \}$ of A with $\displaystyle d(x_k,y_k) \rightarrow 0$

Clearly if $\displaystyle f$ is uniformly continuous the statement is correct but not your proof.
From U.C. $\displaystyle \varepsilon > 0 \Rightarrow \left( {\exists \delta > 0} \right)\left[ {d(x,y) < \delta \Rightarrow p\left( {f(x),f(y)} \right) < \varepsilon } \right]\quad$.
Suppose that $\displaystyle \left( {x_n } \right) \to x\,\& \,\left( {y_n } \right) \to y$ then $\displaystyle \left( {\exists N} \right)\left[ {n \geqslant N \Rightarrow d\left( {x_n ,x} \right) < \frac{\delta }{3} \wedge d\left( {y_n ,y} \right) < \frac{\delta }{3}} \right]$.
Now consider this: If $\displaystyle d(x,y) < \frac{\delta }{3}$ then $\displaystyle d\left( {x_n ,y_n } \right) \leqslant d\left( {x_n ,x} \right) + d\left( {x,y} \right) + d\left( {y,y_n } \right) < \delta \Rightarrow p\left( {f\left( {x_n } \right),f\left( {y_n } \right)} \right) < \varepsilon$.

P.S. I was far too tired to think straight last night.

The converse is easy. If $\displaystyle f$ is not informally continuous then $\displaystyle \left( {\exists \lambda > 0} \right)\left( {\forall \delta > 0} \right)\left( {\exists x_\delta \wedge y_\delta } \right)\left[ {d(x_\delta ,y_\delta ) < \delta \wedge p(f(x_\delta ),f(y_\delta )) \geqslant \lambda } \right]$.
That is the negation of informally continuous.

So $\displaystyle \left( {\forall n} \right)\left( {\exists x_n \wedge y_n } \right)\left[ {d(x_n ,y_n ) < \frac{1}{n} \wedge p(f(x_n ),f(y_n )) \geqslant \lambda } \right]$
That is a contradiction, because $\displaystyle d(x_n ,y_n ) \to 0$.